mathematical induction1.1

(a) Let P(n) be the statement 1 + [1/(1+√2)] + [1/(√2 + √3)] +...+ 1/[√(n-1)+√n] = √n
When n = 1, L.H.S. = 1, R.H.S. = 1
Thus P(1) is true.
Suppose that P(k) is true where k is a positive integer, i.e.
1 + [1/(1+√2)] + [1/(√2 + √3)] +...+ 1/[√(k-1)+√k] = √k
Adding 1/[√k + √(k + 1)] to both sides:
1 + [1/(1+√2)] + [1/(√2 + √3)] +...+ 1/[√(k-1)+√k] + 1/[√k + √(k + 1)] = √k + 1/[√k + √(k + 1)]
= [k + √k√(k + 1) + 1]/[√k + √(k + 1)]
= [(k + 1) + √k√(k + 1)]/[√k + √(k + 1)]
= [√(k + 1)√(k + 1) + √k√(k + 1)]/[√k + √(k + 1)]
= √(k + 1)[√(k + 1) + √k]/[√k + √(k + 1)]
= √(k + 1)
So P(k + 1) is also true. By the principle of MI, P(n) is true for all positive integers n.
(b) Let P(n) be the statement 1x2 - 2x4 + 3x6 - 4x8 +...+ (-1)n-1‧n‧2n = (-1)n-1‧n‧(n+1)
When n = 1, L.H.S. = 2, R.H.S. = 2
Thus P(1) is true.
Suppose that P(k) is true where k is a positive integer, i.e.
1x2 - 2x4 + 3x6 - 4x8 +...+ (-1)k-1‧k‧2k = (-1)k-1‧k‧(k+1)
Adding (-1)k‧(k + 1)‧2(k + 1) to both sides:
1x2 - 2x4 + 3x6 - 4x8 +...+ (-1)k-1‧k‧2k = (-1)k-1‧k‧(k+1) + (-1)k‧(k + 1)‧2(k + 1)
= (-1)k[- k(k + 1) + 2(k + 1)2]
= (-1)k(k + 1)[- k + 2(k + 1)]
= (-1)k(k + 1)(k + 2)
So P(k + 1) is also true. By the principle of MI, P(n) is true for all positive integers n.